Q1.Two dozen eggs and a loaf of bread cost K4.82.Half a dozen eggs and two loaves of bread cost K2.29.Find the cost of a loaf of bread.
2*x+y=4.82
0.5x+2y=2.29
Multiply the second equation (0.5x+2y=2.29) by 4 and take away the x.
2*x+8y=9.16
Subtract 2*x+y=4.82 from 3
7*y=4.34
y=4.34/7=0.62k
The cost would be 0.62k. Hope that helps! :)
Macpherson emily.east new britain matriculation study centre PO BOX 1684,kokopo ENBP.PAPUA NEW GUINEA.
Let's assume the cost of a dozen eggs as 'x' and the cost of a loaf of bread as 'y'.
According to the first statement:
Two dozen eggs (2x) + 1 loaf of bread (y) = K4.82
According to the second statement:
Half a dozen eggs (0.5x) + 2 loaves of bread (2y) = K2.29
We can set up a system of equations with these two statements:
Equation 1: 2x + y = K4.82
Equation 2: 0.5x + 2y = K2.29
To solve the system of equations, we can use the method of substitution or elimination.
Using the method of substitution, we can solve Equation 1 for y and substitute it into Equation 2:
From Equation 1: y = K4.82 - 2x
Substituting y into Equation 2:
0.5x + 2(K4.82 - 2x) = K2.29
Simplifying the equation:
0.5x + 9.64 - 4x = K2.29
-3.5x = K2.29 - 9.64
-3.5x = -7.35
Dividing both sides of the equation by -3.5:
x = K2.10
Now we can substitute the value of x back into Equation 1 to find y:
2(2.10) + y = K4.82
4.20 + y = K4.82
y = K4.82 - 4.20
y = K0.62
Therefore, the cost of a loaf of bread is K0.62.
To find the cost of a loaf of bread, we can assign variables to the unknowns. Let's denote the cost of a loaf of bread as 'B' and the cost of one dozen eggs as 'E'.
From the first statement, we know that two dozen eggs and a loaf of bread cost K4.82. Since there are 12 eggs in a dozen, two dozen eggs would be 24 eggs. Thus, the cost of 24 eggs is 24E.
Using this information, we can create an equation:
24E + B = K4.82 ...(Equation 1)
The second statement states that half a dozen eggs and two loaves of bread cost K2.29. Half a dozen eggs is 6 eggs or 6E.
Similarly, we can create a second equation:
6E + 2B = K2.29 ...(Equation 2)
Now we have a system of two equations with two variables (E and B). Let's solve these equations simultaneously.
Multiplying Equation 2 by 4 to eliminate the E coefficient, we get:
24E + 8B = K9.16 ...(Equation 3)
Subtracting Equation 1 from Equation 3, we eliminate the term with E:
(24E + 8B) - (24E + B) = K9.16 - K4.82
Simplifying, we have:
8B - B = K9.16 - K4.82
7B = K4.34
Dividing both sides by 7, we find:
B = K4.34 / 7
Therefore, the cost of a loaf of bread, B, is approximately K0.62.